Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$, we have $r_1 + r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$.
• Equation of a Circle (Diameter Form): If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a diameter of a circle, the equation of the circle is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
• General Equation of a Circle: $x^2 + y^2 + 2ax + 2by + c = 0$.

Step-by-Step Solution

Step 1: Define the Coordinates and Relate them to the Given Equations Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be the endpoints of the diameter of the circle. The x-coordinates, $x_1$ and $x_2$, are the roots of the equation $x^2 - 4x - 6 = 0$. The y-coordinates, $y_1$ and $y_2$, are the roots of the equation $y^2 + 2y - 7 = 0$. This step sets up the problem by establishing the connection between the coordinates of the points and the given quadratic equations.

Step 2: Apply Vieta's Formulas to Find the Sum and Product of the Roots Using Vieta's formulas for the equation $x^2 - 4x - 6 = 0$, we find the sum and product of the x-coordinates:

• $x_1 + x_2 = -\frac{-4}{1} = 4$
• $x_1 x_2 = \frac{-6}{1} = -6$

Similarly, for the equation $y^2 + 2y - 7 = 0$, we find the sum and product of the y-coordinates:

• $y_1 + y_2 = -\frac{2}{1} = -2$
• $y_1 y_2 = \frac{-7}{1} = -7$ This step utilizes Vieta's formulas to efficiently determine the sum and product of the coordinates, which are necessary for constructing the circle's equation in diameter form.

Step 3: Form the Equation of the Circle Using the Diameter Form Since $PQ$ is a diameter, the equation of the circle is given by: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ Expanding this equation, we get: $x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$ Substituting the values obtained from Vieta's formulas: $x^2 - (4)x + (-6) + y^2 - (-2)y + (-7) = 0$ Simplifying the equation: $x^2 + y^2 - 4x + 2y - 6 - 7 = 0$ $x^2 + y^2 - 4x + 2y - 13 = 0$ This step constructs the specific equation of the circle based on the information derived from the previous steps.

Step 4: Compare the Derived Equation with the General Equation and Find a, b, and c We are given the general equation of the circle as: $x^2 + y^2 + 2ax + 2by + c = 0$ Comparing this with our derived equation $x^2 + y^2 - 4x + 2y - 13 = 0$, we get:

• $2a = -4 \Rightarrow a = -2$
• $2b = 2 \Rightarrow b = 1$
• $c = -13$ This step identifies the values of $a$, $b$, and $c$ by comparing the coefficients of the derived equation with the given general equation.

Step 5: Calculate the Value of a + b - c Now, we substitute the values of $a$, $b$, and $c$ into the expression $a + b - c$: $a + b - c = (-2) + (1) - (-13)$ $a + b - c = -2 + 1 + 13$ $a + b - c = 12$ This final step computes the value of the desired expression using the values of $a, b,$ and $c$ found in the previous steps.

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when applying Vieta's formulas and when comparing coefficients. A common mistake is to forget the negative sign in $-\frac{b}{a}$.
• Diameter Form Formula: Remember the diameter form formula for the circle. It is a direct way to find the equation when the endpoints of a diameter are known.
• Vieta's Formulas: Always remember and correctly apply Vieta's formulas.

Summary

This problem combines the concepts of quadratic equations and circles. By using Vieta's formulas to find the sum and product of the roots of the given quadratic equations and then applying the diameter form of the circle's equation, we were able to find the specific equation of the circle. Comparing this with the general form, we found the values of $a$, $b$, and $c$, and finally calculated $a+b-c$.

The final answer is $\boxed{12}$, which corresponds to option (A).